Similarity and enlargement – Year 9 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. An enlargement multiplies every length by the scale factor: \(6\times2.5=15\) cm (1 mark) and \(4\times2.5=10\) cm (1 mark). Q is 15 cm by 10 cm.
   2. The long side of R corresponds to the long side of P, so the scale factor is \(4.5\div6=\)0.75. (1 mark)
   3. Multiply the short side of P by the same scale factor: \(4\times0.75=\)3 cm. (1 mark)

   A scale factor less than 1 makes the shape smaller, so R is a reduction of P even though it is still called an enlargement in mathematics. Check: in P the sides are in the ratio \(6:4=3:2\); in Q \(15:10=3:2\); in R \(4.5:3=3:2\). Similar rectangles always keep this ratio.

2. Answer to Question 2 (4 marks)

   1. DE corresponds to AB, so the scale factor is \(9\div6=\)1.5. (1 mark)
   2. EF corresponds to BC: \(8\times1.5=\)12 cm (1 mark). DF corresponds to AC: \(7\times1.5=\)10.5 cm (1 mark).
   3. Match the sides by size: smallest with smallest and so on. The ratios are \(12\div6=2\), \(14\div7=2\) and \(16\div8=2\). All three are equal, so yes, the triangle is similar to ABC with scale factor 2. (1 mark)

   The letters tell you which sides correspond: D matches A and E matches B, so DE matches AB. If even one of the three ratios were different, the triangles would not be similar.

3. Answer to Question 3 (4 marks)

   1. With centre (0, 0), multiply each coordinate by the scale factor 3: \((1,1)\to(3,3)\), \((3,1)\to(9,3)\), \((1,2)\to(3,6)\). The new vertices are \((3,3)\), \((9,3)\), \((3,6)\). (2 marks for all three, 1 mark for two)
   2. To undo an enlargement by 3, enlarge by \(\dfrac13\) about the same centre. The transformation is an enlargement (1 mark), scale factor \(\dfrac13\), centre (0, 0) (1 mark).

   A full description of an enlargement needs three things: the word enlargement, the scale factor and the centre. Check (a): the base of T is 2 units long and the image base is \(9-3=6\) units, three times as long.

4. Answer to Question 4 (4 marks)

   1. The sun's rays are parallel, so the stick, the tree and their shadows form similar triangles. The scale factor from the stick's shadow to the tree's shadow is \(13.5\div1.8=7.5\) (1 mark). The tree is 7.5 times as tall as the stick: \(1.2\times7.5=\)9 m (1 mark).
   2. The stick's shadow grows from 1.8 m to 3 m, a factor of \(3\div1.8=\dfrac53\) (1 mark). The tree's shadow grows by the same factor: \(13.5\times\dfrac53=\)22.5 m (1 mark).

   Another way to see it: at any moment, shadow ÷ height is the same for every object. In (a) that ratio is \(1.8\div1.2=1.5\), and \(13.5\div1.5=9\) m. Later it becomes \(3\div1.2=2.5\), and \(9\times2.5=22.5\) m.

5. Answer to Question 5 (4 marks)

   1. Use the same units: 30 m = 3 000 cm. The model is 50 times smaller: \(3000\div50=\)60 cm. (1 mark)
   2. The real building is 50 times larger: \(14\times50=700\) cm, which is 7 m. (1 mark)
   3. Areas are multiplied by the square of the scale factor: \(50^{2}=2\,500\) (1 mark). Real area \(=120\times2500=300\,000\) cm². There are \(100\times100=10\,000\) cm² in 1 m², so this is \(300000\div10000=\)30 m² (1 mark).

   Multiplying the area by 50 only would give 6 000 cm², which is just 0.6 m², far too small for a building.

Total : 20 marks